Sin(2arccos (1/2)+3arctgsqrt3)

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вопрос

First, let's find the values of the trigonometric functions:

Now we substitute these values into the given expression:

2arccos(1/2) = 2 (π/3) = 2π/3
3arctan(sqrt(3)) = 3 (π/3) = π

So, the expression becomes:

sin(2arccos(1/2) + 3arctan(sqrt(3))) = sin(2π/3 + π)

Using the sine addition formula, we have:

sin(2π/3 + π) = sin(2π/3)cos(π) + cos(2π/3)sin(π)

sin(2π/3) = sqrt(3)/2
cos(2π/3) = -1/2
cos(π) = -1
sin(π) = 0

Therefore, the expression simplifies to:

(sqrt(3)/2 -1) + (-1/2 0) = -sqrt(3)/2

Therefore, sin(2arccos(1/2) + 3arctan(sqrt(3))) = -sqrt(3)/2